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Section 6-4 : Surface Integrals of Vector Fields

  1. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = 3x\,\vec i + 2z\,\vec j + \left( {1 - {y^2}} \right)\vec k\) and \(S\) is the portion of \(z = 2 - 3y + {x^2}\) that lies over the triangle in the xy-plane with vertices \(\left( {0,0} \right)\), \(\left( {2,0} \right)\) and \(\left( {2, - 4} \right)\) oriented in the negative \(z\)-axis direction. Solution
  2. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = - x\,\vec i + 2y\,\vec j - z\,\vec k\) and \(S\) is the portion of \(y = 3{x^2} + 3{z^2}\) that lies behind \(y = 6\) oriented in the positive \(y\)-axis direction. Solution
  3. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = {x^2}\,\vec i + 2z\,\vec j - 3y\,\vec k\) and \(S\) is the portion of \({y^2} + {z^2} = 4\) between \(x = 0\) and \(x = 3 - z\) oriented outwards (i.e. away from the \(x\)-axis). Solution
  4. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = \,\vec i + z\,\vec j + 6x\,\vec k\) and \(S\) is the portion of the sphere of radius 3 with \(x \le 0\), \(y \ge 0\) and \(z \ge 0\) oriented inward (i.e. towards the origin). Solution
  5. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = y\,\vec i + 2x\,\vec j + \left( {z - 8} \right)\,\vec k\) and \(S\) is the surface of the solid bounded by \(4x + 2y + z = 8\), \(z = 0\), \(y = 0\) and \(x = 0\) with the positive orientation. Note that all four surfaces of this solid are included in \(S\). Solution
  6. Evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) where \(\vec F = yz\,\vec i + x\,\vec j + 3{y^2}\,\vec k\) and \(S\) is the surface of the solid bounded by \({x^2} + {y^2} = 4\), \(z = x - 3\), and \(z = x + 2\) with the negative orientation. Note that all three surfaces of this solid are included in \(S\). Solution